# Double Intercept Form

Here, we will discuss about the double intercept form of the equation of straight lines.

Let a straight line intersect x-axis and y-axis at $A(a,0)$ and $B(0,b)$ respectively. Then, the intercepts on the x-axis and y-axis are $OA=a$ and $OB=b$ respectively. Also, let $P(x,y)$ be any point on $AB$.

Then, $\text{Slope of AP}=\text{Slope of PB}$ $\frac{y-0}{x-a}=\frac{b-y}{0-x}$ $\frac{y}{x-a}=\frac{b-y}{-x}$ $-xy=bx-xy-ab+ay$ $ab=bx+ay$ $1=\frac{x}{a}+\frac{y}{b}$ $\therefore\frac{x}{a}+\frac{y}{b}=1$

This is the equation of the line whose intercepts are $a$ and $b$ and is known as the double intercept form. This form of the equation of a straight line cannot be used if the line passes through the origin or if it be parallel to any one of the axes.

### Obtain the equation of the straight line passing through the point $(3,4)$ cutting off equal intercepts on the axes.

Here, $\text{x-intercept}=\text{y-intercept}$ $\therefore a=b$ Then the equation of the line is, $\frac{x}{a}+\frac{y}{b}=1$ $\frac{x}{a}+\frac{y}{a}=1$ $x+y=a\text{ __(1)}$

Since this line passes through the point $(3,4)$, $3+4=a$ $\therefore a=7$

Putting the value of $a$ in $\text{(1)}$, $x+y=7$ which is the required equation of the line.

### Find the equation of the straight line which passes through the point $(3,4)$ and makes intercepts on the axes, the sum of whose lengths is $14$.

$\text{Let x-intercept}=a$ $\text{y-intercept}=b$ $\text{Then, }a+b=14$ $b=14-a\text{ __(1)}$

The equation of the line is, $\frac{x}{a}+\frac{y}{b}=1$ $bx+ay=ab$ Since this line passes through $(3,4)$, $3b+4a=ab\text{ __(2)}$

Solving $\text{(1)}$ and $\text{(2)}$, $3(14-a)+4a=a(14-a)$ $42-3a+4a=14a-a^2$ $a^2-13a+42=0$ $(a-6)(a-7)=0$ $\therefore a=6,7$

Taking $a=6$, $b=14-6=8$. Then, the required equation of the line is $\frac{x}{6}+\frac{y}{8}=1$ $\therefore 4x+3y=24$

Taking $a=7$, $b=14-7=7$. Then, the required equation of the line is, $\frac{x}{7}+\frac{y}{7}=1$ $\therefore x+y=7$

### Determine the equation of the line the portion of which, intercepted by the axes, is divided by the point $(-5,4)$ in the ratio $1:2$.

Let the line intersect x-axis and y-axis at $A(a,0)$ and $B(0,b)$ respectively and $AB$ is divided at $C$ in the ratio $1:2$. Then, from section formula for internal division, we have, $-5=\frac{1×0+2×a}{1+2}\text{ and }4=\frac{1×b+2×0}{1+2}$ $-5=\frac{2a}{3}\text{ and }4=\frac{b}{3}$ $\therefore a=\frac{-15}{2}\text{ and }b=12$

Thus, equation of line is, $\frac{x}{a}+\frac{y}{b}=1$ $\frac{-2x}{15}+\frac{y}{12}=1$ $-8x-5y=60$ $\therefore 8x-5y+60=0$